椭圆x^2/a^2+y^2/b^2=1(a>b>0)的左右焦点分别为F1,F2,A是椭圆上的一点,AF1垂直AF2原点O到直线AF1的距离为1/3|OF|1)证明a=根号2 b2)求t属于(0,b)使得下述命题成立:设圆 x^2+y^2=t^2上任意点M(X,Y)处的切线交
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/08 17:49:45
![椭圆x^2/a^2+y^2/b^2=1(a>b>0)的左右焦点分别为F1,F2,A是椭圆上的一点,AF1垂直AF2原点O到直线AF1的距离为1/3|OF|1)证明a=根号2 b2)求t属于(0,b)使得下述命题成立:设圆 x^2+y^2=t^2上任意点M(X,Y)处的切线交](/uploads/image/z/3705728-32-8.jpg?t=%E6%A4%AD%E5%9C%86x%5E2%2Fa%5E2%2By%5E2%2Fb%5E2%3D1%28a%3Eb%3E0%29%E7%9A%84%E5%B7%A6%E5%8F%B3%E7%84%A6%E7%82%B9%E5%88%86%E5%88%AB%E4%B8%BAF1%2CF2%2CA%E6%98%AF%E6%A4%AD%E5%9C%86%E4%B8%8A%E7%9A%84%E4%B8%80%E7%82%B9%2CAF1%E5%9E%82%E7%9B%B4AF2%E5%8E%9F%E7%82%B9O%E5%88%B0%E7%9B%B4%E7%BA%BFAF1%E7%9A%84%E8%B7%9D%E7%A6%BB%E4%B8%BA1%2F3%7COF%7C1%29%E8%AF%81%E6%98%8Ea%3D%E6%A0%B9%E5%8F%B72+b2%29%E6%B1%82t%E5%B1%9E%E4%BA%8E%EF%BC%880%2Cb%29%E4%BD%BF%E5%BE%97%E4%B8%8B%E8%BF%B0%E5%91%BD%E9%A2%98%E6%88%90%E7%AB%8B%EF%BC%9A%E8%AE%BE%E5%9C%86+x%5E2%2By%5E2%3Dt%5E2%E4%B8%8A%E4%BB%BB%E6%84%8F%E7%82%B9M%28X%2CY%29%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E4%BA%A4)
xRMKP++|I I[馐\h5ZLjRB4-MhNn\/8IυWo5g3[hVL6,3NH6E-J}+`p)Ɇu7B )D!3pαd\C|C`;Ρ
CߠRnuTË>zlbR-{}g3Ēp
;prYqS'3XK-